Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

/*
   AtCoder contest code
   coder : yoichidon
 */

public class Main {
    public static String longer(String x, String y){
        if(x.length() > y.length()) return x;
        else return y;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        double[] p = new double[N];
        for(int n=0;n<N;n++) p[n]=sc.nextDouble();

        double[][] dp = new double[N][N+1];
        dp[0][1] = p[0];
        dp[0][0] = 1 - p[0];

        for(int n=1; n<N; n++) for(int head=0; head<=n+1; head++){
            dp[n][head] = dp[n-1][head] * (1-p[n]) + (head==0?0: dp[n-1][head-1] * p[n]);
        }

        double ans = 0;
        for(int head=N; head>N-head; head--) ans += dp[N-1][head];
        System.out.println(ans);

    }
}